Pythagorean triplets

Originally posted by geepamoogle
Very well......
The triplets formulas which help find these quickly and thoroughly are as follows (where a>b).

2*a*b
a^2 - b^2
a^2 + b^2 < hypotenuse

If a and b are relatively prime AND exactly one of the two is even, you'll find a unique triplet.

Have you noticed that one of the triplets
a,b, c is always a multiple of 5?
Why should it be so?
[Here my a, and b are the smaller two of the triplet (a,b,c),
It is not the "a" and the "b" in Geepamoogle's generating formula.]

Originally posted by observantU
I have been able to discover at least 2 more than Geepamoogle...
12-35-37 and 27-36-45...Hence there are at least 18. May be more..

Well, geepamoogle already specified 12-35-37, and 27-36-45 is not independent from 3-4-5. So, no.

Originally posted by howzzat
Have you noticed that one of the triplets
a,b, c is always a multiple of 5?
Why should it be so?

In addition to this, one of the legs is always divisible by 4, and one of the legs is always divisible by 3.

It's in the nature of modulo arithmetic in these bases. Please note that you could have one number divisible by all 3..

Originally posted by LemonJello
Well, geepamoogle already specified 12-35-37, and 27-36-45 is not independent from 3-4-5. So, no.

LemonJello is right. May be there are only 16 of them ..

Originally posted by geepamoogle
In addition to this, one of the legs is always divisible by 4, and one of the legs is always divisible by 3.

It's in the nature of modulo arithmetic in these bases. Please note that you could have one number divisible by all 3..

The question is why should it be so? One of the three triplets being always divisible by either 3, or 4, or 5 indicates that all triplets arise perhaps from the basic triplet (3,4,5). This is my cojecture..May be I am wrong. But the question remains why should it be so?

Originally posted by LemonJello
Well, geepamoogle already specified 12-35-37, and 27-36-45 is not independent from 3-4-5. So, no.

Thanx LemonJ... I concede.. and I withdraw my assertion...

Originally posted by ranjan sinha
The question is why should it be so? One of the three triplets being always divisible by either 3, or 4, or 5 indicates that all triplets arise perhaps from the basic triplet (3,4,5). This is my cojecture..May be I am wrong. But the question remains why should it be so?

Nice conjecture...though. But does it have also something to do with the the nature of modulo arithmetic ? As mentioned by Geepamoogler?

Originally posted by geepamoogle
In addition to this, one of the legs is always divisible by 4, and one of the legs is always divisible by 3.

It's in the nature of modulo arithmetic in these bases. Please note that you could have one number divisible by all 3..

That's remarkable indeed.

One leg or the other always being a multiple of one or more of the three factors 3, or 4 or 5.
This remarkable property is observable for any independent triplet..

[Of course by multiplying each leg of any independent triplet by 60, you can get a triplet, each of whose legs are simultaneously divisible by 3, 4 and 5.].

Originally posted by howzzat
Have you noticed that one of the triplets
a,b, c is always a multiple of 5?
Why should it be so?
[Here my a, and b are the smaller two of the triplet (a,b,c),
It is not the "a" and the "b" in Geepamoogle's generating formula.]

i'll use p and q as the generating numbers (p>q), and let a = 2pq, b = p^2-q^2, and c = p^2 + q^2

a perfect square is always either 0, 1 or 4 (mod5). it also follows that if x^2 = 0 (mod 5), then x=0 (mod 5).

now let's look at the generating formulas for pythagorean triples: if p=0 (mod5) or q=0 (mod5) then clearly 2pq = 0 (mod 5) and so one of the sides is a multiple of 5.

so now assume both of them are NOT 0 (mod5) and we'll see one of the other sides must be 0 (mod 5). there are only 3 cases left:

1. p^2=q^2=1 (mod 5): but then (p^2 - q^2) = 0 (mod5), so b = 0 (mod 5).
2. p^2=q^2=4 (mod 5): by the same logic, p^2-q^2 = b = 0 (mod5).
3. WLOG, p^2=1 (mod5) and q^2=4 (mod5): but then p^2+q^2 = 0 (mod 5) and so c = 0 (mod 5).

a cute fact, with a nice argument by cases in modulo arithmetic. i believe there are similar arguments that show the necessity of a multiple of 3 and a multiple of 4, dealing with the parity of p and q, their relative primality, and the factoring of p^2-q^2 as (p-q)(p+q)... i'm too lazy to try and figure it out though 🙂

For a unique base triplet, it will be generated by a combination of an even and an odd number.

In this case, the 2*p*q term will be divisible by 4.

If you take a look at modulo 3, then square numbers will always be 0(mod3) if the base is divisible by 3 and 1(mod3) if not.

Since 1+1=2 and 2(mod3) is provably not square, the only way to get a triplet (aside from 0+0=0) is 1+0=1.

Hence one of the numbers must always be divisible by 3.

Now, I do believe there are some transformations from one triplet to a higher triplet, but I may be thinking approximations of square roots. More later maybe.

*bleh need a way to delete new posts Keep accidentally hitting the wrong button*

Originally posted by Aetherael
i'll use p and q as the generating numbers (p>q), and let [b]a = 2pq, b = p^2-q^2, and c = p^2 + q^2

a perfect square is always either 0, 1 or 4 (mod5). it also follows that if x^2 = 0 (mod 5), then x=0 (mod 5).

now let's look at the generating formulas for pythagorean triples: if p=0 (mod5) or q=0 (mod5) then clearly 2pq = 0 (mod 5) and so g of p^2-q^2 as (p-q)(p+q)... i'm too lazy to try and figure it out though 🙂[/b]

The conjecture of the author of this thread that all triplets arise from the basic triplet (3,4,5) is indeed true.

Given any primitive (independent pythagorean triplet in its lowest multiple form) triplet (a,b,c) , three new primitive triples can be generated as follows:-
(a1,b1,c1) = (a,b,c)U......................................(1)
(a2,b2,c2) = (a,b,c)A.....................................(2)
(a3,b3,c3) = (a,b,c)D.......................................(3)


Where U A D are defined as the following three matrices respectively : -

U = 1,2,2; ....A =-1,-2,-2;....D = 1,2,2
......2,1,2.......... 2, 1, 2 ..........-2,-1,-2
......2,2,3;.........2, 2, 3;..........2,2,3;


It has been proven that any Pythagorean triplet (p,q,r) can be a primitive Pythagorean triplet iff (p,q,r) = (3,4,5)M;


where M is a finite product of the matrices U , A and D.
In short M can be any matrix from the following pyramidical array of matrices:-
U, .. A, .. D,
UU,UA,UD, .. AU, AA, AD, .. DU, DA, DD,
UUU,UUA,UUD,.. UAU,UAA,UAD,..UDU,UDA,UDD ... AUU,AUA,AUD,etc.
The n-th row in this pyramydical array has 3^n matrices.
Thus all primitive (independent and distinct i.e. in their lowest multiple form) Pythagorean triplets do indeed arise from the basic triplet (3,4,5). You can see the link given below.
http://mathworld.wolfram.com/PythagoreanTriple.html

Originally posted by neverB4chess
The conjecture of the author of this thread that all triplets arise from the basic triplet (3,4,5) is indeed true.

Given any primitive (independent pythagorean triplet in its lowest multiple form) triplet (a,b,c) , three new primitive triples can be generated as follows:-
(a1,b1,c1) = (a,b,c)U................... ...[text shortened]... ,5). You can see the link given below.
http://mathworld.wolfram.com/PythagoreanTriple.html

All this should follow from the triplet generating functions mentioned by some of the contributors to this thread.

Originally posted by ranjan sinha
All this should follow from the triplet generating functions mentioned by some of the contributors to this thread.

What is the final tally then?

Are there only 16 independent triplets below 100?